Express $\partial f/\partial x$ and $\partial f/\partial y$ in terms of $\partial f/\partial r$ and $\partial f/\partial \theta$, and vice versa.

Question

I wrote down the following: \begin{align} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} = \frac{\partial f}{\partial r} \cos(\theta) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (1) \end{align} \begin{align} \frac{\partial f}{\partial x} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x} = - \frac{\partial f}{\partial \theta} \frac{\sin(\theta)}{r} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (2) \end{align} \begin{align} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} = \frac{\partial f}{\partial r} \sin(\theta) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (3) \end{align} \begin{align} \frac{\partial f}{\partial y} = \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial y} = \frac{\partial f}{\partial r} \frac{\cos(\theta)}{r} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (4) \end{align} \begin{align} \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} = \frac{\partial f}{\partial x} \left [-r\sin(\theta) \frac{\partial \theta}{\partial r} + \cos(\theta) \right ] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (5) \end{align} \begin{align} \frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} = \frac{\partial f}{\partial x} \left [ -r \sin(\theta) + \cos(\theta) \frac{\partial r}{\partial \theta}\right ] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (6) \end{align}

It is possible to write $\frac{\partial f}{\partial x}$ purely in terms of $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \theta}$? And same for $\frac{\partial f}{\partial y}$

Also, I think $\partial \theta / \partial r = \partial r / \partial \theta = 0$ so $(5)$ and $(6)$ become: \begin{align} \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} = \frac{\partial f}{\partial x} \left [ \cos(\theta) \right ] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (5') \end{align} \begin{align} \frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} = \frac{\partial f}{\partial x} \left [ -r \sin(\theta) \right ] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (6') \end{align}

Comparing $(5')$ with $(1)$ and comparing $(6')$ with $(2)$ is really confusing to me. Can anyone explain what am I doing incorrect?

Answer 1

Assume that function $f(x, y)$ depends also on $(r, \theta)$ through

$$ \begin{aligned} x &= x(r, \theta) \\ y &= y(r, \theta) \\ \end{aligned} $$

Then, $$ \begin{aligned} \frac{\partial f}{\partial r} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial r} \\ \frac{\partial f}{\partial \theta} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial \theta} \end{aligned} $$

which can be rewritten in a matrix form:

$$ \begin{pmatrix} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} \end{pmatrix} \begin{pmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{pmatrix} $$

from which,

$$ \begin{pmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} \end{pmatrix}^{-1} \begin{pmatrix} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial \theta} \end{pmatrix} $$

Thus, if $$ \begin{aligned} x(r, \theta) &= r\cos{\theta} \\ y(r, \theta) &= r\sin{\theta}, \end{aligned} $$

then

$$ \begin{pmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{pmatrix} = \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -r\sin{\theta} & r\cos{\theta} \end{pmatrix}^{-1} \begin{pmatrix} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial \theta} \end{pmatrix}= \begin{pmatrix} \cos{\theta} & -\frac{\sin{\theta}}{r} \\ \sin{\theta} & \frac{\cos{\theta}}{r} \end{pmatrix} \begin{pmatrix} \frac{\partial f}{\partial r} \\ \frac{\partial f}{\partial \theta} \end{pmatrix} = \begin{pmatrix} \cos{\theta}\frac{\partial f}{\partial r}-\frac{\sin{\theta}}{r}\frac{\partial f}{\partial \theta} \\ \sin{\theta}\frac{\partial f}{\partial r}+\frac{\cos{\theta}}{r}\frac{\partial f}{\partial \theta} \end{pmatrix}\Rightarrow \\ \\ \begin{aligned} \frac{\partial f}{\partial x} &= \cos{\theta}\frac{\partial f}{\partial r}-\frac{\sin{\theta}}{r}\frac{\partial f}{\partial \theta} \\ \frac{\partial f}{\partial y} &= \sin{\theta}\frac{\partial f}{\partial r}+\frac{\cos{\theta}}{r}\frac{\partial f}{\partial \theta} \end{aligned} $$